Question: $f(x) = \begin{cases} 8 & \text{if } x = 1 \\ 3x^{2}-1 & \text{otherwise} \end{cases}$ What is the range of $f(x)$ ?
Solution: First consider the behavior for $x \ne 1$ Consider the range of $3x^{2}$ The range of $x^2$ is $\{\, y \mid y \ge 0 \,\}$ Multiplying by $3$ doesn't change the range. To get $3x^{2}-1$ , we subtract $1$ So the range becomes: $\{\, y \mid y ≥ -1 \,\}$ If $x = 1$, then $f(x) = 8$. Since $8 ≥ -1$, the range is still $\{\, y \mid y ≥ -1 \,\}$.